$\sum\limits_{n=1}^{\infty} \dfrac{n}{1+n^2}$ When applying the integral test, we get a limit that determines whether the series converges or diverges. What is this limit? Choose 1 answer: Choose 1 answer: (Choice A) A $ \lim_{b\to\infty}\left[ b \tan^{-1}(b) - \dfrac{\pi}{4} \right]$ (Choice B) B $\lim_{b\to\infty}\dfrac{b}{b^2+1}$ (Choice C) C $\lim_{b\to\infty} \dfrac{3b}{2b^2+6}$ (Choice D) D $\lim_{b\to\infty} \dfrac{1}{2}\ln\left(\dfrac{1+b^2}{2}\right) $
Explanation: $\dfrac{n}{1+n^2}$ satisfies the conditions for the integral test. This means that $\sum\limits_{n=1}^{\infty} \dfrac{n}{1+n^2}$ converges/diverges together with $\int_1^{\infty} \dfrac{x}{1+x^2} \,dx$. $\int_1^{\infty} \dfrac{x}{1+x^2} \,dx=\lim_{b\to\infty} \dfrac{1}{2}\ln\left(\dfrac{1+b^2}{2}\right) $ In conclusion, the limit that determines whether the series converges or diverges is $\lim_{b\to\infty} \dfrac{1}{2}\ln\left(\dfrac{1+b^2}{2}\right) $.